| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Separable variables - standard (applied/contextual) |
| Difficulty | Standard +0.3 This is a straightforward numerical methods question requiring two Euler method iterations with a simple differential equation, followed by a basic conceptual adjustment to a parameter. The calculations are routine and the reasoning in part (b) is minimal—slightly easier than average for Further Maths FP1. |
| Spec | 1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams4.10a General/particular solutions: of differential equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Identifies \(t_0=0\), \(v_0=0\), \(\left(\frac{dv}{dt}\right)_0=10\) and \(h=0.5\) | B1 | Uses model to identify correct initial conditions and \(h\). May be implied by use in equation |
| \(v_1 = v_0 + h\left(\frac{dv}{dt}\right)_0 \Rightarrow v_1 = 0 + 0.5\times 10 = \ldots\) | M1 | Applies approximation formula with values for \(v_0\), \(\left(\frac{dv}{dt}\right)_0\) and \(h\) |
| \(v_1 = 5\) | A1 | |
| \(\left(\frac{dv}{dt}\right)_1 = -0.1(5)^2 + 10 = \ldots\{7.5\}\); \(v_2 = v_1 + h\left(\frac{dv}{dt}\right)_1 \Rightarrow v_2 = 5 + 0.5\times 7.5 = \ldots\) | M1 | Uses \(v_1\) to find \(\left(\frac{dv}{dt}\right)_1\) and applies approximation formula to find \(v_2\) |
| \(v_2 = 8.75\) so \(8.75\ \text{ms}^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dv}{dt} = -0.1v^2 + A\) where \(0 < A < 10\) | B1 | Reduce the value of 10 (explain it needs reducing). Do not accept 0 or negative values. "Change the 10" is B0 if it does not explain how to change it |
## Question 4:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Identifies $t_0=0$, $v_0=0$, $\left(\frac{dv}{dt}\right)_0=10$ and $h=0.5$ | B1 | Uses model to identify correct initial conditions and $h$. May be implied by use in equation |
| $v_1 = v_0 + h\left(\frac{dv}{dt}\right)_0 \Rightarrow v_1 = 0 + 0.5\times 10 = \ldots$ | M1 | Applies approximation formula with values for $v_0$, $\left(\frac{dv}{dt}\right)_0$ and $h$ |
| $v_1 = 5$ | A1 | |
| $\left(\frac{dv}{dt}\right)_1 = -0.1(5)^2 + 10 = \ldots\{7.5\}$; $v_2 = v_1 + h\left(\frac{dv}{dt}\right)_1 \Rightarrow v_2 = 5 + 0.5\times 7.5 = \ldots$ | M1 | Uses $v_1$ to find $\left(\frac{dv}{dt}\right)_1$ and applies approximation formula to find $v_2$ |
| $v_2 = 8.75$ so $8.75\ \text{ms}^{-1}$ | A1 | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dv}{dt} = -0.1v^2 + A$ where $0 < A < 10$ | B1 | Reduce the value of 10 (explain it needs reducing). Do not accept 0 or negative values. "Change the 10" is B0 if it does not explain how to change it |
---\begin{enumerate}
\item The velocity $v \mathrm {~ms} ^ { - 1 }$, of a raindrop, $t$ seconds after it falls from a cloud, is modelled by the differential equation
\end{enumerate}
$$\frac { \mathrm { d } v } { \mathrm {~d} t } = - 0.1 v ^ { 2 } + 10 \quad t \geqslant 0$$
Initially the raindrop is at rest.\\
(a) Use two iterations of the approximation formula $\left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) _ { n } \approx \frac { y _ { n + 1 } - y _ { n } } { h }$ to estimate the velocity of the raindrop 1 second after it falls from the cloud.
Given that the initial acceleration of the raindrop is found to be smaller than is suggested by the current model,\\
(b) refine the model by changing the value of one constant.
\hfill \mbox{\textit{Edexcel FP1 2022 Q4 [6]}}